Question: Factor completely. $2 x^2 -40 x +200=$
Explanation: First, we take a common factor of $2$. $2 x^2 -40 x +200=2(x^2 -20 x +100)$ Now, let's factor $x^2 -20 x +100$. Both $x^2$ and $100$ are perfect squares, since $x^2=({x})^2$ and $100=({10})^2$. Additionally, $20 x$ is twice the product of the roots of $x^2$ and $100$, since $20 x=2({x})({10})$. $x^2 -20 x +100 = ({x})^2 - 2({x})({10})+({10})^2$ So we can use the square of a difference pattern to factor: ${a}^2 - 2( a)( b)+ {b}^2 =({a} - {b})^2$ In this case, ${a}={x}$ and ${b}={10}$ : $ ({x})^2 - 2({x})({10})+({10})^2 =({x} -{10})^2$ $\begin{aligned} 2 x^2 -40 x +200 &=2(x^2 -20 x +100) \\\\ &=2(x -10)^2 \end{aligned}$ In conclusion, the complete factorization is $2(x -10)^2$ Remember that you can always check your factorization by expanding it.